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The first and third of these equations provide a

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Unformatted text preview: for the system (and for each of the blocks individually). (b) Applying Eq. 5- 2 to block 1, we find T1 = m1a = (12.0kg ) 0.970m/s2 = 11.6N. ( ) (c) In order to find T2, we can either analyze the forces on block 3 or we can treat blocks 1 and 2 as a system and examine its forces. We choose the latter. T2 = ( m1 + m2 ) a = (12.0 kg + 24.0 kg ) 0.970 m/s 2 = 34.9 N . ( ) 5P57. The free- body diagram for each block is shown below. T is the tension in the cord and θ = 30° is the angle of the incline. For block 1, we take the +x direction to be up r the incline and the +y direction to be in the direction of the normal force FN that the plane exerts on the block. For block 2, we take the +y direction to be down. In this way, the accelerations of the two blocks can be represented by the same symbol a, without ambiguity. Applying Newton’s second law to the x and y axes for block 1 and to the y axis of block 2, we obtain T − m1 g sin θ = m1a FN − m1 g cos θ = 0 m2 g − T = m2 a 3 PHY2048 Spring 2014 HW3 respectively. The first and third of these equations provide a simultaneous set for obtaining values of a and T. The second equation is not needed in this problem, since the normal force is neither asked for nor is it needed as part of some further computation (such as can occur in formulas for friction). (a) We add the first and third equations above: m2g – m1g sin θ = m1a + m2a. Consequently, we find ( m2 − m1 sin θ ) g = [2.30 kg − (3.70 kg)sin 30.0°] 9.8...
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